Question:
If an emitter current is changed by $4 \mathrm{~mA}$, the collector current changes by $3.5 \mathrm{~mA}$.
The value of $\beta$
Correct Option: 1
Solution:
(1)
Given :
$\Delta \mathrm{I}_{\mathrm{E}}=4 \mathrm{~mA}$
$\Delta \mathrm{I}_{\mathrm{C}}=3.5 \mathrm{~mA}$
we know that, $\alpha=\frac{\Delta \mathrm{I}_{\mathrm{c}}}{\Delta \mathrm{I}_{\mathrm{E}}}$
$\Rightarrow \alpha=\frac{3.5}{4}=\frac{7}{8}$
Also, $\beta=\frac{\alpha}{1-\alpha}$, so
$\beta=\frac{\frac{7}{8}}{1-\frac{7}{8}}=\frac{7}{1}$
$\beta=7$
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